Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define rep(i, l, r) for(int i = (l), i##end = (r);i <= i##end;++i)
using std::cin; using std::cout;
const int maxn = 200200;
typedef long long ll;
int a[maxn], b[maxn], n, k;
ll dp[maxn], pa[maxn];
int l[maxn], r[maxn];
std::map<int,int> map;
typedef std::map<int, int>::iterator iter;
inline iter split(int r) { iter x = map.lower_bound(r); return x -> first == r ? x : map.emplace_hint(x, r, std::prev(x) -> second); }
inline void set(int l,int r,int c) { if(l <= r) { iter R = split(r + 1), L = split(l); map.erase(L, R), map[l] = c; } }
inline ll get(int x) { int c = split(x) -> second; return c ? dp[c] + (x + k - r[c]) % k : 0; }
int main() {
	std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> k, map[0] = map[k] = 0;
	rep(i, 1, n) {
		cin >> a[i] >> b[i];
		pa[i] = pa[i - 1] + a[i];
		if(b[i] == 1) {
			if(a[i] + a[i] > k) return cout << -1 << '\n', 0;
			l[i] = (ll) pa[i - 1] % k * (k - 2) % k, r[i] = (ll) pa[i - 0] % k * ll(k - 2) % k;
			dp[i] = get(r[i]);
			if(l[i] <= r[i]) set(r[i] + 1, k - 1, i), set(0, l[i] - 1, i);
			else set(r[i] + 1, l[i] - 1, i);
		}
	}
	ll ans = 1e18;
	rep(i, 1, n) ans = std::min(ans, get(l[i]));
	cout << ans + pa[n] + pa[n] << '\n';
}